Question

In the LR circuit shown (consisting an ideal inductor, 3 ideal batteries and 2 resistors), what is the variation of current $I$ as a function of time? The switch is closed at time $t=0$ with inductor not carrying any current.

Answer 1

I(t) = \frac{8V}{R} (1 - e^{-Rt/(2L)})

Answer 2

1. Interpret the Circuit: The most plausible interpretation of the given LR circuit diagram is that the inductor L is in series with the 2V battery. This combination is then in series with a parallel combination of two branches.

   • Branch 1: Top resistor R and V battery.
   • Branch 2: Bottom resistor R, 3V battery, and the switch. The current $I$ is the current flowing through the inductor L and the 2V battery.

2. Simplify the Parallel Part: Calculate the equivalent resistance ($R_{eq,p}$) and equivalent voltage ($V_{eq,p}$) of the two parallel branches.

   • For resistance: The two resistors R are in parallel, so $R_{eq,p} = \frac{R \times R}{R + R} = \frac{R}{2}$.
   • For voltage: Using the formula for parallel voltage sources (assuming positive terminals are aligned to contribute positively to the voltage from left to right): $V_{eq,p} = \frac{V_1/R_1 + V_2/R_2}{1/R_1 + 1/R_2} = \frac{V/R + 3V/R}{1/R + 1/R} = \frac{(V+3V)/R}{2/R} = \frac{4V/R}{2/R} = 2V$.

3. Form the Equivalent Series Circuit: The entire circuit simplifies to a series combination of the inductor L, the 2V battery, the equivalent parallel resistance $R_{eq,p}$, and the equivalent parallel voltage $V_{eq,p}$.

   • Total effective voltage ($V_{eff}$): The 2V battery and the $V_{eq,p}$ (2V) are in series. Assuming they are aiding each other (as per common circuit problem conventions for net voltage calculation), $V_{eff} = 2V + 2V = 4V$.
   • Total effective resistance ($R_{eff}$): The only resistance in the equivalent series circuit is $R_{eq,p} = R/2$. So, $R_{eff} = R/2$.

4. Apply KVL to the Equivalent LR Circuit: The differential equation for current in an LR circuit is $L \frac{dI}{dt} + I R_{eff} = V_{eff}$. Substituting the values: $L \frac{dI}{dt} + I \left(\frac{R}{2}\right) = 4V$.

5. Solve the Differential Equation: The solution for current in an LR circuit when the switch is closed at $t=0$ and initial current is zero is $I(t) = I_{ss} (1 - e^{-t/\tau})$.

   • Steady-state current ($I_{ss}$): At steady state ($t \to \infty$), the inductor acts as a short circuit. So, $I_{ss} = \frac{V_{eff}}{R_{eff}} = \frac{4V}{R/2} = \frac{8V}{R}$.
   • Time constant ($\tau$): $\tau = \frac{L}{R_{eff}} = \frac{L}{R/2} = \frac{2L}{R}$.

6. Final Expression for Current: Substitute $I_{ss}$ and $\tau$ into the general solution: $I(t) = \frac{8V}{R} (1 - e^{-t/(2L/R)}) = \frac{8V}{R} (1 - e^{-Rt/(2L)})$.

The initial condition $I(0)=0$ is satisfied by this solution.

Answer: The variation of current $I$ as a function of time is given by: $I(t) = \frac{8V}{R} \left(1 - e^{-Rt/(2L)}\right)$